Pointwise strong laws need not give a uniform law

Fix a finite set $A\subseteq [0,1]$. If $A=\mathrm{\varnothing }$, then $P(A)=0$. Otherwise write $m=\mathrm{\mid }A\mathrm{\mid }\ge 1$. Given $ϵ>0$, for each $x\in A$ let $I_x=\{y\in [0,1]:\mathrm{\mid }y-x\mathrm{\mid }<ϵ\mathrm{/}(2m)\}$. These $m$ intervals cover $A$, and each has length at most $ϵ\mathrm{/}m$. By [A1] and finite subadditivity [F2], $0\le P(A)\le {\sum }_{x\in A}P(I_x)\le ϵ$. Since this holds for every $ϵ>0$, $P(A)=0$. By [F1], each event $\{X_i\in A\}$ has probability zero; countable subadditivity [F2] gives $P(\mathrm{\exists }i\ge 1:X_i\in A)=0$. Hence with probability one $P_n(A)=0=P(A)$ for every $n$. This proves the pointwise strong law for this fixed $A$, without asserting one common event for all finite sets.

For any realized finite sample $x_1,\dots ,x_n$, the set $A_n=\{x_1,\dots ,x_n\}$ is finite, so $A_n\in \mathcal{A}$, $P_n(A_n)=1$, and $P(A_n)=0$. Every indicator deviation is at most one, so the supremum over $\mathcal{A}$ equals one for every sample. This supremum is the constant measurable function one; by [F4] it does not converge to zero in probability.